Question

In dreptunghiul ABCD (AB>BC) se considera AE perpendicular pe BD, E∈BD. Se stie ca DE= 2radical din 2 si EB= 4 radical din 2. a)Calculati perimetrul dreptunghiului ABCD. b)Calculati raportul dintre ariile: ΔAED si ΔAEB.

Answer 1

bd=be+ed=4√2+2√2=6√2
ae²=be×ed⇔ae²=4√2×2√2=4×2×2=8×2=16⇒ae=√16⇒ae=4
in Δaed: ae²+ed²=ad²
4²+(2√2)²=ad²
16+√8²=ad²
16+8=ad²
24=ad²⇒ad=√24⇒ad=2√6
in Δaeb
ae²+be²=ab²
4²+(4√2)²=ab²
16+√32²=ab²
16+32=ab²
48=ab²⇒ab=√48⇒ab=4√3
abcd dreptunghi⇒ad=bc=4√3 si ab=cd=2√6
perimetrul=2×(4√3+2√6)=8√3+4√6⇔4×(2√3+√6)
in Δaed fie ex perpendicular ad unde x apartine ad⇒ex×ad=ae×ed
ex×2√6=4×2√2
ex×2√6=8√2
ex=8√2/2√6⇒ex=4√2/√6⇔ex=4√2×√6/6⇔ex=4√12/6⇒ex=2√12/3
arie aed=ad×ex/2=2√6×2√12/3 totul supra 2=4√72/3/2=4√72/3×1/2=4√72/6=4×6√2/6=4√2
in Δaeb fie ey perpendicular ab unde y apartine ab
ey×ab=ae×eb
ey×4√3=4×4√2
ey×4√3=16√2
ey=16√2/4√3=4√2/√3=4√2×√3/3=4√6/3
arie aeb=ey×ab/2=4√6/3×4√3totul supra 2=16√18/3/2=16√18/3×1/2=16√18/6=16×3√2/6=48√2/6=8√2
deci arie aed=4√2 si arie aeb=8√2⇒arie aeb=2×arie aed